Let n = 0 n 3 ( ( 2 n ) ) + ( 2 n - 1 ) ( n ) ( n ) ( ( 2 n ) ) = a e + b e + c , where a , b , c and e = n = 0 1 n . Then a 2 - b + c is equal to ________ . [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let $\sum_{n = 0}^{\infty} \frac{n^{3} ((2 n)!) + (2 n - 1) (n!)}{(n!) ((2 n)!)} = a e + \frac{b}{e} + c,$ where $a, b, c \in ℤ$ and $e = \sum_{n = 0}^{\infty} \frac{1}{n!}$ . Then $a^{2} - b + c$ is equal to ________ . [2023]

Ans.
(26)

We have, $\sum_{n = 0}^{\infty} \frac{n^{3} (2 n!) + (2 n - 1) (n!)}{(n!) ((2 n)!)} = a e + \frac{b}{e} + c$

$\Rightarrow \sum_{n = 0}^{\infty} \frac{1}{(n - 3)!} + \sum_{n = 0}^{\infty} \frac{3}{(n - 2)!} + \sum_{n = 0}^{\infty} \frac{1}{(n - 1)!} + \sum_{n = 0}^{\infty} \frac{1}{(2 n - 1)!} - \sum_{n = 0}^{\infty} \frac{1}{(2 n)!} = a e + \frac{b}{e} + c$

$\Rightarrow e + 3 e + e + \frac{1}{2} (e - \frac{1}{e}) - \frac{1}{2} (e + \frac{1}{e}) = a e + \frac{b}{e} + c$

$\Rightarrow 5 e - \frac{1}{e} = a e + \frac{b}{e} + c$

$Comparing, we get, a = 5, b = - 1, c = 0$

$a^{2} - b + c = 25 + 1 = 26$

Want Us to Email you About Special Offers & Updates?